Population Growth and Compound depreciation
According to Lexmark," the population of the world remains constant that means the total number of animals including social animals in each year is same. But the social animals are increased year by year and rare animals and wild animals are decreased year by year". So, the number of people are increased yearly. In case of Nepal, the population of 2058 and 2064 are different. The population of 2064 BS is more than that of 2058 BS. The growth of the population indicates the additional population than the previous year. The growth number of the population does not remain constant. So, it is calculated in compounded way.
Let 'P' be the population at the beginning of certain year and P_{T }be the population after 'T' years, then
\(P_T = P \left(1 + \frac {R_1} {100}\right)\left(1 + \frac {R_2} {100}\right)\left(1 + \frac {R_3} {100}\right) ................... \left(1 + \frac {R_T} {100}\right)\)
where \(R_1, R_2, R_3, ....... R_T \) are the rates of 1^{st}, 2^{nd}, 3^{rd}, ............ t^{th} years respectively.
If R_{1}=R_{2}= R_{3}=............., R_{T }= R then \(P_T = P \left(1 + \frac {R} {100}\right)^T\)
\(P_T = P \left(1 + \frac {R} {100}\right)^T\)
In case of decrease in population, \(P_T = P \left(1  \frac {R} {100}\right)^T\).
Compound depreciation
The value of the machine is decreased yearly due to various reasons like wear and tear, inefficiency etc. The reduced value of the machine is called the depreciated value. The amount of depression is different in successive years even the rate of depreciation is same. Such depression is called compound depreciation. Therefore, the formula for depreciated value or scrap value which is used to solve the problems of compound depreciation can be written as:
\(P_T = P \left (1  \frac {R} {100} \right ) ^ T\)
\(or, S = V\left (1  \frac {R} {100} \right ) ^ T\)
Where S, V, R, T indicates amount after compound depreciation, principal (original value), rate of depreciation in percentage per annum and number of years for which goods are used respectively. The amount of compound depreciation determined by subtracting scrap value 'S' from the original value V i.e. Compound depreciation = V  S or P  P_{T}.
When the number of periods is not an integer, compound interest for the integral period is calculated first and the simple interest at the given rate for the fractional period is calculated:
a. Simple interest and compound interest of the first period are equal.
b. The compound interest of more than one simple interest.
c. The difference of amounts of two consecutive periods is equal to the interest on the amount of first conversion period.
d. The difference of compound interest of the two consecutive periods is equal to the interest of the first conversion period.
Solution:
Present value (P_{T}) = Rs 14580
Previous value (P) = Rs 18000
Time (T) = 2 years
Depreciate rate (R) = ?
\begin{align*} P_T &= P \left( 1  \frac{R}{100} \right)^T \\ or, 14580 &= 18000 \left( 1  \frac{R}{100} \right)^2\\ or, 0.81 &=\left( 1  \frac{R}{100} \right)^2\\ or, (0.9)^2 &= \left(1  \frac{R}{100}\right)^2\\ or, 0.9 &= 1  \frac{R}{100}\\ or, \frac{R}{100} &= 1 0.9 \\ or, R&= 0.1\times100\\ \therefore R &= 10\% \:\:\:_{Ans.} \end{align*}
Solution:
The price before 2 years (P) = Rs 6000
Present price (P_{T}) = Rs 5415
Time(T) = 2 years
Depreciation Rate (R) = ?
\begin{align*} P_T &= P \left( 1  \frac{R}{100} \right)^T \\ or, 5415 &= 6000 \left( 1  \frac{R}{100} \right)^2\\ or, 0.9025 &=\left( 1  \frac{R}{100} \right)^2\\ or, (0.95)^2 &= \left(1  \frac{R}{100}\right)^2\\ or, 0.95 &= 1  \frac{R}{100}\\ or, \frac{R}{100} &= 1 0.95 \\ or, R&= 0.05\times100\\ \therefore R &= 5\% \:\:\:_{Ans.} \end{align*}
Solution:
The population of a village one year ago (P) = 10,000
Present population (P_T) = 10210
Growth rate (R) = ?
Time (T) = 1 year
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^2 \\ or, 10210 &= 10000 \left(1 + \frac{R}{100}\right)^T \\ or, \frac{10210}{10000} &= 1 + \frac{R}{100}\\ or, 1.021 &= 1 + \frac{R}{100}\\ or, 1.021  1 &= \frac{R}{100}\\ or, R &= 0.021 \times 100\% \\ \therefore R &= 2.1\% \end{align*}
Solution:
Population of the last year = 7200
Here,
\begin{align*} Population \: after \: one \: year &= 7200  7200 \times \frac{5}{100}  7200 \times \frac{2}{100}\\ &= 7200  360 144 \\ &= 7200  504\\ &= 6996 \:\:\: _{Ans.} \end{align*}
Solution:
The population of the village last year = 2000
\begin{align*} \text{Population of a place after one year } &= 2000 + 2000 \times \frac{3}{100} + 2000 \times \frac{2}{100}\\ &= 2000+60+40\\ &= 2100 \:\:\:\:\: _{Ans} \end{align*}
Solution:
Present population (P_{T}) = 242000
Growth rate (R) = 10%
Time (T) = 2 years
Population of a town 2 years ago (P) = ?
Now,
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 242000 &= P \left( 1 + \frac{10}{100}\right)^2\\ 242000 &= P(1.1)^2 \\ or, P &= \frac{242000}{1.21} \\ \therefore P &= 200,000 \end{align*}
Solution:
The present population (P) = 64000
Growth rate (R) = 5%
Time (T) = 2 years
Population of a town after 2 years (P_{T}) = ?
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ P_T &= 64000 \left( 1 + \frac{1}{100}\right)^2\\ &= 64000 (1.05)^2\\ &=64000 \times 1.1025 \\ &= 70560 \:\:\:\: _{Ans.} \end{align*}
Solution:
Present price of motor cycle (P) = Rs 140,000
Rate of depreciation (R) = 7%
Time (T) = 2 years after
The price of motorcycle 2 years (P_{T}) = ?
\begin{align*} P_T &= P \left( 1  \frac{R}{100}\right)^T \\ &= 140000 \left( 1  \frac{7}{100}\right)^2 \\ &= 140000[1  0.07]^2 \\ &= 140000 \times (0.93)^2\\ &= 140000 \times 0.8649 \\ &= Rs \: 121086 \: \: \: _{Ans.} \end{align*}
Solution:
Present population (P_{T}) = 40000
Growth rate (R) = 2% + 3% =5%
Time (T) = 2 yrs
Population of the town after 2 years (P_{2})= ?
Now,
\begin{align*} P_2 &= P_t \left( 1 + \frac{R}{100} \right)^T \\ &= 40000 \left( 1 + \frac{5}{100} \right)^2\\ &=40000\left( \frac{105}{100} \right)^2\\ &= 40000 \times \frac{105 \times 105}{100 \times 100} \\ &= 44,100 _{Ans.} \end{align*}
Solution:
Present Population (P) = 96000
Growth rate (R) = 5%
population after T year (P_{T}) = 105840
Time (T) = ?
\begin{align*} P_T &= P \left( 1 + \frac{R}{100} \right)^T \\ 105840 &= 96000\left( 1 + \frac{5}{100} \right)^T\\ or, \frac{105840}{96000} &= \left( \frac{105}{100}\right)^T \\ or, 1.1025 &= (1.05)^T \\ or, (1.05)^2 &= (1.05)^T\\ T &= 2 \end{align*}
\(\therefore \) Time = 2 years
Solution:
Present population (P_{T}) = 500000
Time (T) = 3 years ago
Population of a town 3 years ago = P
We know that,
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 500000 &= P\left( 1 + \frac{3}{100}\right)^3 \\ or, 500000&= P(1.03)^3\\ or, P &= \frac{500000}{1.092727}\\P&= 457570.83\end{align*}
The population before 3 years = 457571 (Approx)
After one year ago
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 500000 &= P\left( 1 + \frac{3}{100}\right)^1\\ or, P \times 1.03 &= 500000\\ or, P &= \frac{500000}{1.03}\\ P&=515000 \end{align*}
The population before 1 year = 515000 \(_{Ans.}\)
Solution:
The population of a village before 2 years (P) = 25000
Time (T) = 2 years
Rate of growth (R) = 3%
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\&= 25000 \left( 1 + \frac{3}{100}\right)^2\\ &= 25000 \times (1.03)^3\\ &= 25000 \times 1.0609\\ &= 26522.5 \approx 26522 \end{align*}
No. of died people = 500
Population after 2 years = 26522  500 = 26022
Population before 1 year (P) =26022
Time (T) =1 year
Population on growth rate (R) = 3%
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ &=26022 \left( 1 + \frac{3}{100}\right)^1 \\ &=26022 \times \frac{103}{100}\\ &= 26803 \:(Approx)\end{align*}
\(\therefore \) The present population of village = 26803 \(\:\:_{Ans}\)
Solution:
Total population = 30000
Number of people by migration = 5800
\(\therefore\) Present population (P_{T}) = 30000  5800 = 24200
Growth rate (R) = 10%
Time (T) = 2 years
We know that,
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ or, 24200&= P \left( 1 + \frac{10}{100}\right)^2 \\ or, 24200 &= P(1.1)^2 \\ or, P \times 1.21 &= 24200 \\ or, P &= \frac{24200}{1.21} \\ \therefore P &= 20000 \end{align*}
\(\therefore\) The population of the town at first = 20,000 \(\:\:_{Ans.}\)
Solution:
The price of a computer before 2 years (P) = Rs 44100
Present price of computer (P_{T}) = Rs 40000
Time (T) = 2 years
Rate of depreciation (R) = ?
We know that,
\begin{align*} P_T &= P \left( 1  \frac{R}{100}\right)^T\\ or, 40000 &= 44100 \left( 1  \frac{R}{100}\right)^2\\ or, \frac{40000}{44100} &= \left( 1  \frac{R}{100}\right)^2\\ or, \left( 1  \frac{R}{100}\right)^2 &= \left(\frac{20}{21}\right)^2\\ or, 1  \frac{R}{100}&= \frac{20}{21}\\ or,1\frac{20}{21} &= \frac{R}{100} \\ or, \frac{2120}{21} &= \frac{R}{100}\\ or, \frac{R}{100} &= \frac{1}{21}\\ or, R &= \frac{100}{21}\\ R &= 4.76\% \end{align*}
\(\therefore\) the rate of depreciation = 4.76% \(\:\:\: _{Ans.}\)
Solution:
\begin{align*} \text{Price of the one ropani of land 3 year ago (P)} &= \frac{1250000}{4}\\ & = Rs\: 312500 \end{align*}
Price of one ropaniof land at present (P_{T}) = Rs 160000
Rate of depreciated (R) = ?
Time (T) = 3 years
We know that,
\begin{align*} P_T &= P \left( 1  \frac{R}{100}\right)^T\\ or, 160000 &= 3125000 \left( 1  \frac{R}{100}\right)^3\\ or, \frac{160000}{312500}&= \left( 1  \frac{R}{100}\right)^3\\ or, \frac{64}{125}&= \left( 1  \frac{R}{100}\right)^3\\ or, \left( \frac{4}{5} \right)^3 &= \left( 1  \frac{R}{100}\right)^3\\ \frac{4}{5}&=1\frac{R}{100}\\ or, \frac{R}{100}&= 1\frac{4}{5} \\ or, R &= \frac{54}{5} \times 100\\ R &=20\% \end{align*}
The depreciated rate (R) = 20%
Solution:
Present population (P_{T}) = 24895 + 320 = 25215
Time (T) = 2 years
Growth rate (R) = 2.5%
Population before 2 years (P) =?
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ or, 25215 &= P \left( 1 + \frac{2.5}{100}\right)^2\\or, 25215 &= P(1.021)^2\\ or, P \times 1.050625 &= 25215\\ P &= \frac{25215}{1.050625}\\ P &= 24000 \end{align*}
\(\therefore\) the population of a village 2 years before = 24000 \(\:_{Ans.}\)
Solution:
The population of a town in 2070 (P) = 1,00,000
Population growth (R) = 2%
Time (T) = 1 year
The population of a town in 2071 (P_{T}) = ?
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ &=100000\left( 1 + \frac{2}{100}\right)^1\\ &= 100000 \times \frac{102}{100}\\ &= 1,02,000 \end{align*}
Migrated people = 8,000
The population of a town after migrated people in 2071 (P) = 102000 + 8000 =1,10,000
Time (T) = 2 years
Rate (R) = 2%
The population of a town in the beginning of 2071 (P_{T}) =?
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ &=110000\left( 1 + \frac{2}{100}\right)^2\\ &= 110000 \times \frac{102}{100} \times \frac{102}{100}\\ &= 1,14,444 \end{align*}
\(\therefore\) The population of a town in the beginning of 2071 = 1,14,444 \(\:\:_{Ans.}\)

The population of a city at present is 1,70,000 and it grows at the rate of 2% per year What will be the population after 2 years?
166868
157878
176868
186868

2 years ago, the population of a town was 120000. If the rate of growth of population is 3% pa what is the present population of the town?
127208
1227312
117308
127308

The population of place was 2000. Within a year the population is increased 3% by birthrate and 2% by migration. How much population is there now?
2200
2150
2000
2100

The present population of a town is 40,000.If the population increases by 2% by birth and 3% by immigration, what will be the population of the town after 2 years?
43200
42100
44100
43100

The population of a village was 10000 one year ago.The population at present is 10210.Find the population growth rate.
3%
2%
2.1%
5%

The population of a village increase every year by 5%. At the end of two years, the total population of the village was 10000, if 1025 was migrated to other places, what was the population of the village in the beginning?
10000
12570
9000
11000

The population of a village increases every year by 5%. At the end of two years, the total population of the village was 26000.If 460 people were migrated to the other village what was the population of the village in the beginning?
23000
21000
22000
24000

The present population of Namche bazar is 50000 and the annual growth rate of this village is 4%.What will be its population after two years ?
54080
50000
528040
514080

The population of Salyantar was 30000 two years ago. If the population at present is 33075, find the population growth rate.
3.5 %
2.2 %
5%
4.5%

The present population of Ghandrung village is 13310 and the rate of growth is 10% .How many year before the population was 10000 ?
2 Years
3Years
4 years
10 months

The population of a town is 1622400. Two years ago the population was 1500000.Find the population of the town 1 year at the same rate of growth.
1269548
1687294
1687296
1479463

3 years ago, the population of a village was 25000. The rate of growth of population is 3% one year ago, 500 people died of epidemic diseases, what is the present population of the village?
27804
26904
26807
26804

The population of a village in 2062 BS and 2063 BS was 133,100 respectively. What was the population of the village in 2060 BS if the growth rate is the same?
110000
120000
130000
100000

Three years ago, there were 1000 students in a secondary school. Since the past three years the rule, "a group of 5 student should bring one new student for enrollment" was imposed to increase the number of student. What is the numbers of students at present?
1730
1645
1727
1728

The population of a village has increased from 20000 to 22050 at the rate of 50 per thousand per year. Find the time period.
4 years
1 year
2 years
3 years

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